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AQA Core 4 - Thursday 12th June 2014 (PM) - Official Thread

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Reply 40

chloetheuincorn

No matter how many times I do Vector questions and get help for Vectors from tutors I still can't do it. :frown:

Well, it's only the ones that tell you the shape and find a point. I can never visualize what I need. :mad:

Reply 41

Photovoltaic

My least favourite module, I find it extremely easy to forget the content for some reason and can't keep it in my memory so I'm not overly confident

Reply 42

a123a

This exam is the maths exam I'm dreading the most :s-smilie:

Reply 43

winniestar57

Can someone please help me with b part (ii)?

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Reply 44

pirategyal

Original post by winniestar57

Can someone please help me with b part (ii)?

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Sorry its sideways :/ does it make sense?

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Reply 45

winniestar57

Original post by pirategyal

Sorry its sideways :/ does it make sense?

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Thanks - but I still don't understand. What method are you using to differentiate?

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Reply 46

pirategyal

Original post by winniestar57

Thanks - but I still don't understand. What method are you using to differentiate?

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Chain rule, if y=e^(kx), dy/dx=ke^(kx)
Its just something we have to remember

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Reply 47

winniestar57

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Reply 48

pirategyal

Original post by winniestar57

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Yeah but you lost a minus when you differentiated, its -60e^(-0.25t) times -0.25

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(edited 9 years ago)

Reply 49

winniestar57

Original post by pirategyal

Yeah but you lost a minus when you differentiated, its -60e^(-0.25t) times -0.25

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Okay, thank you :smile:

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Reply 50

Nick214

What are the hardest c4 papers (not solomons) do people think?

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Reply 51

niceguy95

how do you integrate:

1/4x

i forogt

Reply 52

Math12345

Original post by niceguy95

how do you integrate:

1/4x

i forogt

∫1/(4x) dx = ∫1/4 *1/x dx = 1/4∫1/x dx= 1/4 lnx +c

Reply 53

niceguy95

How do you integrate 1/square root (1+2y)

Reply 54

Math12345

Original post by niceguy95

How do you integrate 1/square root (1+2y)

∫ (1+2y)^-1/2 dy

Now use inspection...

Reply 55

niceguy95

Original post by Math12345

∫ (1+2y)^-1/2 dy

Now use inspection...

inspection.......? :confused:

Reply 56

Math12345

Original post by niceguy95

inspection.......? :confused:

∫ (1+2y)^-1/2 dy
=1/(1/2 *2) * (1+2y)^1/2 +c
=(1+2y)^1/2 +c

∫(ax+b)^n dx = 1/[(n+1)*a] *(ax+b)^(n+1) +c
1/(new power*derivative of bracket) * (function)^new power +c

Only works if the derivative is a constant.

(edited 9 years ago)

Reply 57

vethopefully

guys i've completely for got how to do implicit differentiation?
Can anyone write down the steps/method for it?

e.g curve :9x^2-6xy+4x^2=3 , find the coordinates of the two stationary points

Reply 58

winniestar57

Original post by vethopefully

guys i've completely for got how to do implicit differentiation?
Can anyone write down the steps/method for it?

e.g curve :9x^2-6xy+4x^2=3 , find the coordinates of the two stationary points

This is what I would do, correct if I am wrong :smile:

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Reply 59

than4711

Hi guys, anybody knows which side in the diffrential equation to put (+c)
In this mark scheme, the equation was dx/dt, so i think the +c must be put to the side with t but it's not :eek: